How Does an Electric Field Affect Spring Motion on Another Planet?

[~christina~]

Homework Statement

On a certain planet, the acceleration dute to gravity is the same as that on Earth but there is a strong downward electrical field with the field being uniform close to the planet's surface. The magnitude of this electric field is E= 4.75x10^5N/C. A 1.00kg mass carrying a charge of 100\mu C is connected to a vertical spring for which the spring constant is 200N/m. The spring is not stretched when the mass is released from rest.

a) by what maximum amount does the spring expand?
b) What is the equillibrium postition of the mass
c) show that the mass executes simple harmonic motion

Homework Equations

The Attempt at a Solution

E= 4.75x10^5 N/C
m= 100kg
q= 100\muC
k= 200N/m
g= 9.8m/s^2

hm..I would think that the downward gravitational field would make the force pushing a person or mass down would be greater.

sort of lost here.

Can someone help? Thanks

I was thinking that well there is force downward on the spring that is initially unstretched but this Fy is composed of electric field and gravity so:

E= F/q
I think I can find the force since the E is given and the q is given right?

Then would I just add that to the gravitational force?

and then use that on spring as in F=-kx since I would have the force ?

 

[Nabeshin]

What's the issue? You have all your steps laid out, why don't you do it? Have a little faith in yourself, especially when you have no problems in your logic.

P.S: q is not 200. Might just be a mis-type but that would be one huge charge ^^

 

[~christina~]

What's the issue? You have all your steps laid out, why don't you do it? Have a little faith in yourself, especially when you have no problems in your logic.

P.S: q is not 200. Might just be a mis-type but that would be one huge charge ^^

I said k was 200...never said q was 200.

see the thing is that it's not really the first step that is the problem usually it's the 2nd and and 3rd step that have me like this

 

[Nabeshin]

Mistake, I meant 100. In your "attempt at solution" it says q=100, which should be 100*10^-6. Oh and you wrote m=100kg, which should be 1kg lol.

 

[~christina~]

q= 100x10^-6 C

a and

m= 1.00kg

alright so...if I did the problem like I said I would then I would get:

E=F/q

F=Eq

F= (4.75x10^5 N/C) (1.00x10^-6C)= .475N

and the force due to gravity equals

F=mg

F= 1.00kg(9.8m/s^2)= 9.8N

F_{total}= .475N + 9.8N= 10.275N

and that would be put into

F=-kx

10.275 N/(200N/m)= x

x=5.1375x10^-2 m

but as to finding 2 is it using the fact that when a object is at equillibrium that v=0?

Thanks

 

[Nabeshin]

Actually, that's the equilibrium position you found. Equilibrium is defined as when the sum of the forces and torques is equal to zero. However, because of good old Newton's first, the object will continue moving through this point. Consider the work the forces (downward, constant forces) and spring (changing force) will do and the point at which v=0 is the maximum displacement.

 

[Doc Al]

q= 100x10^-6 C

a and

m= 1.00kg

alright so...if I did the problem like I said I would then I would get:

E=F/q

F=Eq

F= (4.75x10^5 N/C) (1.00x10^-6C)= .475N

Check that value for charge.

and the force due to gravity equals

F=mg

F= 1.00kg(9.8m/s^2)= 9.8N

OK.

F_{total}= .475N + 9.8N= 10.275N

and that would be put into

F=-kx

10.275 N/(200N/m)= x

x=5.1375x10^-2 m

but as to finding 2 is it using the fact that when a object is at equillibrium that v=0?

As Nabeshin points out, here you have calculated the equilibrium position (measured from the starting point), which is the point where the net force is zero.

Use energy conservation to solve for the lowest point--that's the point where v = 0.

 

[~christina~]

fixed

q= 100x10^-6 C

a and

m= 1.00kg

alright so...if I did the problem like I said I would then I would get:

E=F/q

F=Eq

F= (4.75x10^5 N/C) (100x10^-6C)= 47.5N

and the force due to gravity equals

F=mg

F= 1.00kg(9.8m/s^2)= 9.8N

F_{total}= 47.5N + 9.8N= 57.3

and that would be put into

F=-kx

57.3N/(200N/m)= x

x=2.865x10^-1 m

__________________
a)

for the conservation of energy I'd say that...at max displacement
PE= 1/2kA^2
KE= 0
x= A

initial it's

PE_i + KE_i + PE_si= KE_f + PE_f + PE_sf

mgh + 0 + 0= 1/2mv^2 + 0 + 1/2kx^2

right or not?

 

[DavidWhitbeck]

Don't forget electric potential energy.

 

[~christina~]

Don't forget electric potential energy.

PE_i + KE_i + PE_si+ PE_ei= KE_f + PE_f + PE_sf+ PE_ei

mgh + 0 + 0 + q_o \Delta V = 1/2mv^2 + 0 + 1/2kx^2

is that right? not sure how the equation for potential Electrical Energy would look if in the conservation of energy equation.

Thanks

 

[DavidWhitbeck]

The electric field is uniform, but you're using the PE associated with the field of a point mass.

 

[~christina~]

The electric field is uniform, but you're using the PE associated with the field of a point mass.

so..is it fine now?

mgh + 0 + 0 -q_o Ed = 1/2mv^2 + 0 + 1/2kx^2 + 0

Thanks

 

[Doc Al]

mgh + 0 + 0 -q_o Ed = 1/2mv^2 + 0 + 1/2kx^2 + 0

I assume you're solving part a? So you're comparing the energy at the point of release with the energy at the lowest point.

(1) Why is the electrical PE negative? The electric force points downward, just like gravity.
(2) Why is there a KE term on the right? At the lowest point, the speed is zero. (If it wasn't zero, that means it would still be moving down and thus not yet at the lowest point.)

Be sure to express the distances in terms of the same variable. I see h, d, and x: all you need is one.

 

[~christina~]

I assume you're solving part a? So you're comparing the energy at the point of release with the energy at the lowest point.

(1) Why is the electrical PE negative? The electric force points downward, just like gravity.
(2) Why is there a KE term on the right? At the lowest point, the speed is zero. (If it wasn't zero, that means it would still be moving down and thus not yet at the lowest point.)

Be sure to express the distances in terms of the same variable. I see h, d, and x: all you need is one.

okay, trying to fix this based on what you said.

mgd + q_o Ed + 1/2mv^2 = 1/2kd^2

about the electric field: how would I determine if it is possitive or negative?

 

[Doc Al]

okay, trying to fix this based on what you said.

mgd + q_o Ed + 1/2mv^2 = 1/2kd^2

Almost there: It's released from rest.

about the electric field: how would I determine if it is possitive or negative?

You are told that the field is downward and the charge is positive (I assume), so the electric force acts downward just like gravity.

 

[~christina~]

Almost there: It's released from rest.

hm...

mgd + q_o Ed = 1/2kd^2

You are told that the field is downward and the charge is positive (I assume), so the electric force acts downward just like gravity.

so if the charge is negative, would that just make the sign negative since it would go up instead of downwards?

 

[~christina~]

okay I don't know how to show that something executes simple harmonic motion.

need help on that last part.

Thank you

 

[Doc Al]

mgd + q_o Ed = 1/2kd^2

Good. Now you can solve for d.

so if the charge is negative, would that just make the sign negative since it would go up instead of downwards?

Right.

okay I don't know how to show that something executes simple harmonic motion.

Show that the restoring force is proportional to displacement from equilibrium. (Like a spring obeying Hooke's law: F = -kx.)

 

[~christina~]

Good. Now you can solve for d.

mgd + q_o Ed = 1/2kd^2

um...well I seem to have issues with finding the d but I got this far..

(1.00kg)(9.8m/s^2)d +(100x10^-6C)(4.75x10^5 N/C)d= 1/2(200N/m)(d)^2
9.8Nd + 47.5Nd= 100N/m d^2

100N/m d^2 - 57.3d= 0

(I know I factor but I'm not sure what number to factor with for this...for example if it is
2x^2-4x= 0 then the equation would be factored to 2x(x-2)= 0 and thus 2x=0 or x-2= 0 and answer would be 2)

Show that the restoring force is proportional to displacement from equilibrium. (Like a spring obeying Hooke's law: F = -kx.)

"restoring force"

wouldn't that be hooke's law, though?
(didn't really get it in class and can't find it online)

I'm being told to draw it but for a exam I don't think it would be acceptable to draw.

Thanks.

 

[~christina~]

anyone?

 

[Doc Al]

mgd + q_o Ed = 1/2kd^2

um...well I seem to have issues with finding the d but I got this far..

(1.00kg)(9.8m/s^2)d +(100x10^-6C)(4.75x10^5 N/C)d= 1/2(200N/m)(d)^2
9.8Nd + 47.5Nd= 100N/m d^2

100N/m d^2 - 57.3d= 0

(I know I factor but I'm not sure what number to factor with for this...for example if it is
2x^2-4x= 0 then the equation would be factored to 2x(x-2)= 0 and thus 2x=0 or x-2= 0 and answer would be 2)

Hint: Just divide all terms by d.

"restoring force"

wouldn't that be hooke's law, though?
(didn't really get it in class and can't find it online)

I'm being told to draw it but for a exam I don't think it would be acceptable to draw.

Thanks.

Write the force as a function of vertical displacement. Then show it has a form that matches Hooke's law: F = -kx.

 

[Nabeshin]

In your equation you can factor out a d so either d=0 or the other thing.

To show that the restoring force is proportional to displacement, just derive a force equation, and if it has an x term in it, that's shown.

 

[~christina~]

Hint: Just divide all terms by d.

alright...got confused there.

so...
100N/m d^2 - 57.3d= 0

d(100d-57.3)= 0
d=0
d=0.573m

Write the force as a function of vertical displacement. Then show it has a form that matches Hooke's law: F = -kx.

I know that there's the F=mg (gravitational force on the object)
then it has the Spring force F=-kx which is directed in the opposite direction from the gravitational force.

I think those are the only forces but there is also the electric field force, which is

\sum Fy= mg-kx

and I'm not sure now what to do.

Thanks

 

[Doc Al]

alright...got confused there.

so...
100N/m d^2 - 57.3d= 0

d(100d-57.3)= 0
d=0
d=0.573m

Good.

I know that there's the F=mg (gravitational force on the object)
then it has the Spring force F=-kx which is directed in the opposite direction from the gravitational force.

I think those are the only forces but there is also the electric field force, which is

\sum Fy= mg-kx

and I'm not sure now what to do.

Thanks

Add in the electric force to get the total force on the mass. You need to show that this force is proportional to the displacement from equilibrium and oppositely directed. Hints: Write the total force expression. Change variables to write the position as a function of displacement from equilibrium (call it y, if you like). You found the equilibrium position in part b; use it. (Use it symbolically, not with numbers.)